Question

If $\alpha=\log _{ e } \sqrt{\cot \frac{\pi}{12}}$, then find the value of $\left(\frac{\displaystyle\sum_{k=0}^{\infty} e^{-2 k \alpha}}{\displaystyle\sum_{k=0}^{\infty}(-1)^k e^{-2 k x}}\right)^2$

Answer 1

Expression $=\frac{1+ e ^{-2 \alpha}+ e ^{-4 \alpha}+\ldots \ldots \infty}{1- e ^{-2 \alpha}+ e ^{-4 \alpha}+\ldots \ldots \infty}=\frac{1+ e ^{-2 \alpha}}{1- e ^{-2 \alpha}}=\left(\frac{ e ^{2 \alpha}+1}{ e ^{2 \alpha}-1}\right)$
As, $e ^\alpha=\sqrt{\cot \frac{\pi}{12}} \Rightarrow e ^{2 \alpha}=(2+\sqrt{3})$.
$\therefore$ Expression $=\sqrt{3}$.